MATH 113, SUMMER 2005, HOMEWORK 2 SOLUTION
SOLUTIONS AUTHORED BY JOHN B. FRALEIGH
EDITED BY BENJAMIN JOHNSON
Due June 27
Assignment:
section 3; 1,2,6,10,12,15,18,21,26,34
section 4; 1,3,5,9,12,13,16,17,19,21,25,32,40,41
section 5; 7,8,10,12,13,16,18,19,26,27,29,35,39,43,45,53,54,55
Section 3
1. i) φmust be one to one.
ii) φ[S] must be all of S0.
iii) φ(a∗b)=φ(a)∗0φ(b) for all a,b∈S.
2. It is an isomorphism; φis one to one, onto, and φ(n+m)=−(n+m)=(−n)+(−m)=φ(n)+φ(m)
for all m,n∈Z.
6. It is not an isomorphism because φdoes not map Qonto Q.φ(a),−1 for all a∈Q.
10. It is an isomorphism. For any base a,1, the exponential function f(x)=axmaps Rone to one
onto R+, and φis the exponential map with a=0.5. We have φ(r+s)=0.5(r+s)=(0.5r)(0.5s)=
φ(r)φ(s).
12. It is not an isomorphism because φis not one to one: φ(sin x)=cos 0 =1 and φ(x)=1.
15. It is not an isomorphism because φdoes not map Fonto F. Note that φ(f)(0) =0f(0) =0. Thus
there is no element of Fthat is mapped by φinto the constant function 1.
18. a. For φto be an isomorphism, we must have a∗b=φa+1
3∗φb+1
3=φa+1
3+b+1
3=φa+b+2
3=
a+b+1. The identity element is φ(0) =−1.
b. Using the fact that φ−1must also be an isomorphism, we must have a∗b=φ−1(3a−1) ∗
φ−1(3b−1) =φ−1((3a−1) +(3b−1)) =φ−1(3a+3b−2) =a+b−1
3. The identity element
is φ−1(0) =1/3.
21. The definition is incorrect. It should be stated that hS,∗iand hS0,∗0iare binary structures, φ
must be one to one and onto S0, and the universal quantifier for all a,b∈Sshould appear in an
appropriate place.
Let hS,∗iand hS0,∗0ibe binary structures. A map φ:S→S0is an isomorphism if and only if
φis one to one and onto S0, and φ(a∗b)=φ(a)∗0φ(b) for all a,b∈S.
26. One-to-one: Suppose that φ−1(a0)=φ−1(b0) for a0,b0∈S0. Then a0=φ(φ−1(a0)) =φ(φ−1(b0)) =b0,
so φ−1is one to one.
Onto: Let a∈S. Then φ−1(φ(a)) =a, so φ−1maps S0onto S.
Homomorphism property: Let a0,b0∈S0. Now φ(φ−1(a0∗0b0)) =a0∗0b0.
Because φis an isomorphism, φ(φ−1(a0)∗φ−1(b0)) =φ(φ−1(a0)) ∗0φ(φ−1(b0)) =a0∗0b0also.
Because φis one to one, we conclude that φ−1(a0∗b0)=φ−1(a0)∗0φ−1(b0).
34. Let the set be {a,b}. We need to decide whether interchanging the names of the letters everywhere
in the table and then writing the table again in the order afirst and bsecond gives the same table
or a different table. The same table is obtained if and only if in the body of the table, diagonally
Date: June 27, 2005.
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